What is the solution for 9y^{''}-64y=0,y(0)=1,y^'(0)=0 ?

The solution for 9y^{''}-64y=0,y(0)=1,y^'(0)=0 is y=(1-1/2)e^{(8t)/3}+1/2 e^{-(8t)/3}

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9y^{''}-64y=0,y(0)=1,y^'(0)=0

9 y^{^{''}} - 64 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y = (1 - \frac{1}{2}) e^{\frac{8 t}{3}} + \frac{1}{2} e^{- \frac{8 t}{3}}

Solution steps

9 y^{''} (t) - 64 y = 0

Solve linear ODE:

y = (1 - \frac{1}{2}) e^{\frac{8 t}{3}} + \frac{1}{2} e^{- \frac{8 t}{3}}

y = (1 - \frac{1}{2}) e^{\frac{8 t}{3}} + \frac{1}{2} e^{- \frac{8 t}{3}}

5 y^{^{''}} + 21 y^{^{'}} + 4 y = 0, y (0) = 1, y^{^{'}} (0) = - 19.2 y

4 y^{^{''}} - y^{^{'}} - 3 y = 0, y (0) = 9, y^{^{'}} (0) = 0.25

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y (\frac{1}{3}) = 9, y (0) = 3

5 y^{^{''}} + 80 y = 0

y^{^{''}} + 7 y^{^{'}} + 10 y = 0, y (0) = 11, y^{^{'}} (0) = - 49

What is the solution for 9y^{''}-64y=0,y(0)=1,y^'(0)=0 ?
The solution for 9y^{''}-64y=0,y(0)=1,y^'(0)=0 is y=(1-1/2)e^{(8t)/3}+1/2 e^{-(8t)/3}